Simplify the following expression: $y = \dfrac{8x^2- 41x+36}{x - 4}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(8)}{(36)} &=& 288 \\ {a} + {b} &=& &=& {-41} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $288$ and add them together. The factors that add up to ${-41}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-9}$ and ${b}$ is ${-32}$ $ \begin{eqnarray} {ab} &=& ({-9})({-32}) &=& 288 \\ {a} + {b} &=& {-9} + {-32} &=& -41 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({8}x^2 {-9}x) + ({-32}x +{36}) $ Factor out the common factors: $ x(8x - 9) - 4(8x - 9)$ Now factor out $(8x - 9)$ $ (8x - 9)(x - 4)$ The original expression can therefore be written: $ \dfrac{(8x - 9)(x - 4)}{x - 4}$ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ This leaves us with $8x - 9; x \neq 4$.